Integrand size = 24, antiderivative size = 104 \[ \int \frac {x^3}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\frac {b x^4 \sqrt {c+d x^8}}{8 a (b c-a d) \left (a+b x^8\right )}+\frac {(b c-2 a d) \arctan \left (\frac {\sqrt {b c-a d} x^4}{\sqrt {a} \sqrt {c+d x^8}}\right )}{8 a^{3/2} (b c-a d)^{3/2}} \] Output:
1/8*b*x^4*(d*x^8+c)^(1/2)/a/(-a*d+b*c)/(b*x^8+a)+1/8*(-2*a*d+b*c)*arctan(( -a*d+b*c)^(1/2)*x^4/a^(1/2)/(d*x^8+c)^(1/2))/a^(3/2)/(-a*d+b*c)^(3/2)
Time = 1.35 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.19 \[ \int \frac {x^3}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=-\frac {b x^4 \sqrt {c+d x^8}}{8 a (-b c+a d) \left (a+b x^8\right )}+\frac {(b c-2 a d) \arctan \left (\frac {a \sqrt {d}+b \sqrt {d} x^8+b x^4 \sqrt {c+d x^8}}{\sqrt {a} \sqrt {b c-a d}}\right )}{8 a^{3/2} (b c-a d)^{3/2}} \] Input:
Integrate[x^3/((a + b*x^8)^2*Sqrt[c + d*x^8]),x]
Output:
-1/8*(b*x^4*Sqrt[c + d*x^8])/(a*(-(b*c) + a*d)*(a + b*x^8)) + ((b*c - 2*a* d)*ArcTan[(a*Sqrt[d] + b*Sqrt[d]*x^8 + b*x^4*Sqrt[c + d*x^8])/(Sqrt[a]*Sqr t[b*c - a*d])])/(8*a^(3/2)*(b*c - a*d)^(3/2))
Time = 0.40 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {965, 296, 291, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^3}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx\) |
\(\Big \downarrow \) 965 |
\(\displaystyle \frac {1}{4} \int \frac {1}{\left (b x^8+a\right )^2 \sqrt {d x^8+c}}dx^4\) |
\(\Big \downarrow \) 296 |
\(\displaystyle \frac {1}{4} \left (\frac {(b c-2 a d) \int \frac {1}{\left (b x^8+a\right ) \sqrt {d x^8+c}}dx^4}{2 a (b c-a d)}+\frac {b x^4 \sqrt {c+d x^8}}{2 a \left (a+b x^8\right ) (b c-a d)}\right )\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {1}{4} \left (\frac {(b c-2 a d) \int \frac {1}{a-(a d-b c) x^8}d\frac {x^4}{\sqrt {d x^8+c}}}{2 a (b c-a d)}+\frac {b x^4 \sqrt {c+d x^8}}{2 a \left (a+b x^8\right ) (b c-a d)}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {1}{4} \left (\frac {(b c-2 a d) \arctan \left (\frac {x^4 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^8}}\right )}{2 a^{3/2} (b c-a d)^{3/2}}+\frac {b x^4 \sqrt {c+d x^8}}{2 a \left (a+b x^8\right ) (b c-a d)}\right )\) |
Input:
Int[x^3/((a + b*x^8)^2*Sqrt[c + d*x^8]),x]
Output:
((b*x^4*Sqrt[c + d*x^8])/(2*a*(b*c - a*d)*(a + b*x^8)) + ((b*c - 2*a*d)*Ar cTan[(Sqrt[b*c - a*d]*x^4)/(Sqrt[a]*Sqrt[c + d*x^8])])/(2*a^(3/2)*(b*c - a *d)^(3/2)))/4
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d)) Int[ (a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] || !LtQ[q, -1 ]) && NeQ[p, -1]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /; Free Q[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]
Time = 15.42 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.87
method | result | size |
pseudoelliptic | \(\frac {-\frac {\sqrt {d \,x^{8}+c}\, b \,x^{4}}{b \,x^{8}+a}+\frac {\left (2 a d -c b \right ) \operatorname {arctanh}\left (\frac {a \sqrt {d \,x^{8}+c}}{x^{4} \sqrt {a \left (a d -c b \right )}}\right )}{\sqrt {a \left (a d -c b \right )}}}{8 a \left (a d -c b \right )}\) | \(90\) |
Input:
int(x^3/(b*x^8+a)^2/(d*x^8+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/8/a/(a*d-b*c)*(-(d*x^8+c)^(1/2)*b*x^4/(b*x^8+a)+(2*a*d-b*c)/(a*(a*d-b*c) )^(1/2)*arctanh(a*(d*x^8+c)^(1/2)/x^4/(a*(a*d-b*c))^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (88) = 176\).
Time = 0.22 (sec) , antiderivative size = 467, normalized size of antiderivative = 4.49 \[ \int \frac {x^3}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\left [\frac {4 \, \sqrt {d x^{8} + c} {\left (a b^{2} c - a^{2} b d\right )} x^{4} - {\left ({\left (b^{2} c - 2 \, a b d\right )} x^{8} + a b c - 2 \, a^{2} d\right )} \sqrt {-a b c + a^{2} d} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{16} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{8} + a^{2} c^{2} - 4 \, {\left ({\left (b c - 2 \, a d\right )} x^{12} - a c x^{4}\right )} \sqrt {d x^{8} + c} \sqrt {-a b c + a^{2} d}}{b^{2} x^{16} + 2 \, a b x^{8} + a^{2}}\right )}{32 \, {\left ({\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} x^{8} + a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}\right )}}, \frac {2 \, \sqrt {d x^{8} + c} {\left (a b^{2} c - a^{2} b d\right )} x^{4} + {\left ({\left (b^{2} c - 2 \, a b d\right )} x^{8} + a b c - 2 \, a^{2} d\right )} \sqrt {a b c - a^{2} d} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{8} - a c\right )} \sqrt {d x^{8} + c} \sqrt {a b c - a^{2} d}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{12} + {\left (a b c^{2} - a^{2} c d\right )} x^{4}\right )}}\right )}{16 \, {\left ({\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} x^{8} + a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}\right )}}\right ] \] Input:
integrate(x^3/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="fricas")
Output:
[1/32*(4*sqrt(d*x^8 + c)*(a*b^2*c - a^2*b*d)*x^4 - ((b^2*c - 2*a*b*d)*x^8 + a*b*c - 2*a^2*d)*sqrt(-a*b*c + a^2*d)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2* d^2)*x^16 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^8 + a^2*c^2 - 4*((b*c - 2*a*d)*x^1 2 - a*c*x^4)*sqrt(d*x^8 + c)*sqrt(-a*b*c + a^2*d))/(b^2*x^16 + 2*a*b*x^8 + a^2)))/((a^2*b^3*c^2 - 2*a^3*b^2*c*d + a^4*b*d^2)*x^8 + a^3*b^2*c^2 - 2*a ^4*b*c*d + a^5*d^2), 1/16*(2*sqrt(d*x^8 + c)*(a*b^2*c - a^2*b*d)*x^4 + ((b ^2*c - 2*a*b*d)*x^8 + a*b*c - 2*a^2*d)*sqrt(a*b*c - a^2*d)*arctan(1/2*((b* c - 2*a*d)*x^8 - a*c)*sqrt(d*x^8 + c)*sqrt(a*b*c - a^2*d)/((a*b*c*d - a^2* d^2)*x^12 + (a*b*c^2 - a^2*c*d)*x^4)))/((a^2*b^3*c^2 - 2*a^3*b^2*c*d + a^4 *b*d^2)*x^8 + a^3*b^2*c^2 - 2*a^4*b*c*d + a^5*d^2)]
\[ \int \frac {x^3}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {x^{3}}{\left (a + b x^{8}\right )^{2} \sqrt {c + d x^{8}}}\, dx \] Input:
integrate(x**3/(b*x**8+a)**2/(d*x**8+c)**(1/2),x)
Output:
Integral(x**3/((a + b*x**8)**2*sqrt(c + d*x**8)), x)
\[ \int \frac {x^3}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int { \frac {x^{3}}{{\left (b x^{8} + a\right )}^{2} \sqrt {d x^{8} + c}} \,d x } \] Input:
integrate(x^3/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="maxima")
Output:
integrate(x^3/((b*x^8 + a)^2*sqrt(d*x^8 + c)), x)
Leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (88) = 176\).
Time = 0.18 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.28 \[ \int \frac {x^3}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=-\frac {1}{8} \, d^{\frac {3}{2}} {\left (\frac {{\left (b c - 2 \, a d\right )} \arctan \left (\frac {{\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{{\left (a b c d - a^{2} d^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left ({\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b c - 2 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a d - b c^{2}\right )}}{{\left ({\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a d + b c^{2}\right )} {\left (a b c d - a^{2} d^{2}\right )}}\right )} \] Input:
integrate(x^3/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="giac")
Output:
-1/8*d^(3/2)*((b*c - 2*a*d)*arctan(1/2*((sqrt(d)*x^4 - sqrt(d*x^8 + c))^2* b - b*c + 2*a*d)/sqrt(a*b*c*d - a^2*d^2))/(a*b*c*d - a^2*d^2)^(3/2) + 2*(( sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*b*c - 2*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2 *a*d - b*c^2)/(((sqrt(d)*x^4 - sqrt(d*x^8 + c))^4*b - 2*(sqrt(d)*x^4 - sqr t(d*x^8 + c))^2*b*c + 4*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*a*d + b*c^2)*(a* b*c*d - a^2*d^2)))
Timed out. \[ \int \frac {x^3}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {x^3}{{\left (b\,x^8+a\right )}^2\,\sqrt {d\,x^8+c}} \,d x \] Input:
int(x^3/((a + b*x^8)^2*(c + d*x^8)^(1/2)),x)
Output:
int(x^3/((a + b*x^8)^2*(c + d*x^8)^(1/2)), x)
\[ \int \frac {x^3}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {\sqrt {d \,x^{8}+c}\, x^{3}}{b^{2} d \,x^{24}+2 a b d \,x^{16}+b^{2} c \,x^{16}+a^{2} d \,x^{8}+2 a b c \,x^{8}+a^{2} c}d x \] Input:
int(x^3/(b*x^8+a)^2/(d*x^8+c)^(1/2),x)
Output:
int((sqrt(c + d*x**8)*x**3)/(a**2*c + a**2*d*x**8 + 2*a*b*c*x**8 + 2*a*b*d *x**16 + b**2*c*x**16 + b**2*d*x**24),x)